Problem 6

Question

In Problems, find a vector equation for the line through the given points. $$ (3,2,1),\left(\frac{5}{2}, 1,-2\right) $$

Step-by-Step Solution

Verified

Answer

The vector equation is $ \vec{r}(t) = \langle 3 - \frac{1}{2}t, 2 - t, 1 - 3t \rangle $.

1Step 1: Write the Position Vectors

Firstly, identify the position vectors of the two given points. The position vector for the point $(3, 2, 1)$ is $\vec{A} = \langle 3, 2, 1 \rangle$ and for $\left(\frac{5}{2}, 1, -2\right)$ is $\vec{B} = \left\langle \frac{5}{2}, 1, -2 \right\rangle$.

2Step 2: Find the Direction Vector

Calculate the direction vector $\vec{d}$ by subtracting $\vec{A}$ from $\vec{B}$:\[ \vec{d} = \vec{B} - \vec{A} = \left\langle \frac{5}{2}, 1, -2 \right\rangle - \langle 3, 2, 1 \rangle \]Perform the subtraction:\[ \vec{d} = \left\langle \frac{5}{2} - 3, 1 - 2, -2 - 1 \right\rangle = \left\langle -\frac{1}{2}, -1, -3 \right\rangle \]

3Step 3: Write the Vector Equation of the Line

The vector equation of a line through a point $\vec{A}$ in the direction of $\vec{d}$ is given by:\[ \vec{r}(t) = \vec{A} + t\vec{d} \]Substituting $\vec{A} = \langle 3, 2, 1 \rangle$ and $\vec{d} = \left\langle -\frac{1}{2}, -1, -3 \right\rangle$, we get:\[ \vec{r}(t) = \langle 3, 2, 1 \rangle + t\left\langle -\frac{1}{2}, -1, -3 \right\rangle \]

4Step 4: Simplify the Vector Equation

Simplify the expression to find the parametric form of the line:\[ \vec{r}(t) = \langle 3 - \frac{1}{2}t, 2 - t, 1 - 3t \rangle \]This vector equation represents the line through the given points.

Key Concepts

Position VectorDirection VectorParametric FormVector Subtraction

Position Vector

A position vector is a fundamental concept in vector mathematics. It represents a point in space relative to an origin. This is especially crucial when dealing with vector equations of lines. The position vector is determined by the coordinates of the point. For instance, if a point is given as

$(3, 2, 1)$
$\left(\frac{5}{2}, 1, -2\right)$

the corresponding position vectors would be

$\vec{A} = \langle 3, 2, 1 \rangle$
$\vec{B} = \left\langle \frac{5}{2}, 1, -2 \right\rangle$

These vectors describe the location of the points with respect to the origin of the coordinate system.

Direction Vector

The direction vector is a crucial element of the vector equation of a line. It informs us about the direction in which the line extends. To determine the direction vector, we need to perform subtraction between two position vectors. This is where vector subtraction comes into play.
For our example with points

$\vec{A} = \langle 3, 2, 1 \rangle$
$\vec{B} = \left\langle \frac{5}{2}, 1, -2 \right\rangle$

the direction vector, $\vec{d}$, is computed as:

$\vec{d} = \vec{B} - \vec{A}$

This renders the direction vector:

$\vec{d} = \left\langle -\frac{1}{2}, -1, -3 \right\rangle$

This vector essentially points from one given point to the other, determining the line's trajectory.

Parametric Form

Expressing a line in parametric form provides a way to describe all points along the line with one equation. The parametric form of a line uses a position vector and a direction vector, alongside a parameter $t$. This parameter varies over the real numbers to produce all possible points on the line.
Simply put, a line's parametric equation in three-dimensional space is:

$\vec{r}(t) = \vec{A} + t\vec{d}$

Here, $\vec{A}$ is a point on the line, and $\vec{d}$ is the direction vector. In our example, substituting

$\vec{A} = \langle 3, 2, 1 \rangle$
$\vec{d} = \left\langle -\frac{1}{2}, -1, -3 \right\rangle$

we get the parametric form as:

$\vec{r}(t) = \langle 3 - \frac{1}{2}t, 2 - t, 1 - 3t \rangle$

This format is powerful for determining any point along the line by simply choosing a value for $t$.

Vector Subtraction

Vector subtraction is a necessary operation when dealing with vectors, especially in finding direction vectors. To perform vector subtraction, you subtract the corresponding components of one vector from another.
Given two vectors: